Statistics Homework Solutions 166

Statistics Homework Solutions 166 - 166 CHAPTER 6 0 versus...

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Unformatted text preview: 166 CHAPTER 6 0 versus H1 : median of X median of Y ¡ median of Y ¡ ¡ The test statistic W is the sum of the ranks corresponding to the X sample. 281 5. The sample sizes are m 15 and n 15. ¡ ¡ ¨ W ¡ Since n and m are both greater than 8, compute the z-score of W and use the z table. mm n 1 2 mn m n 1 12 2 03. £ ¥ ¡ ¢ £ ¥ ¢ ¤ ¢ ¡ ¢ W ¨ ∆, the P-value is the sum of the areas ¡ ¨ 0 0424. ¨ ¡ 0 0212 ¡ ¨ ¢ 0 0212 ¨ ¨ ¡ From the z table, P median of Y ¡ Since the alternate hypothesis is of the form median of X to the right of z 2 03 and to the left of z 2 03. ¡ ¡ ¤ z ¡ The P-value is fairly small, and so it provides reasonably strong evidence that the population medians are different. (c) No, the X sample is heavily skewed to the right, while the Y sample is strongly bimodal. It does not seem reasonable to assume that these samples came from populations of the same shape. 19. (a) Let µA be the mean thrust/weight ratio for Fuel A, and let µB be the mean thrust/weight ratio for Fuel B. The appropriate null and alternate hypotheses are H0 : µA µB 0 versus H1 : µA µB 0. ¡ ¡ ¡ ¨ 53 019, sB ¨ ¡ 16, B 2 7294, nB ¡ ¡ 2 5522, nA ¨ ¨ ¡ 54 919, sA 16. ¡ ¨ ¡ (b) A The number of degrees of freedom is 2 2 72942 16 2 72942 16 16 1 ¢ 2 ¥ £ ¨ ¤ 29, rounded down to the nearest integer. ¢ ¡ µB 0. ∆, the P-value is the area to the right of t ¡ ¡ ©£¥ 0 05. A computer package gives P 2 0339. 0 026. ¨ ¡ ¨ ¨ ¡ ¡ ¨ ¤ ¡ ¨ ¡ ¡ ¨ 0 05, we can conclude at the 5% level that the means are different. ¨ ¨ Since P P µB 0 versus H1 : µA ¡ £ From the t table, 0 025 ¨ ¢ ¡ ¡ ¤ Since the alternate hypothesis is of the form µA µB ¨ ¨ ¨ The null and alternate hypotheses are H0 : µA 2 0339. ¡ 2 72942 16 £ 2 55222 16 ¡ 0 ¨ £ ¡ 53 019 ¡ ¥ 54 919 ¡ ¨ t29 ¨ ν ¡ 2 55222 16 2 55222 16 2 16 1 21. (a) Yes. (b) The conclusion is not justified. The engineer is concluding that H0 is true because the test failed to reject. 65, O 2 ¡ ¡ ¡ ¡ £ j O , as shown in the following table. Oi O ¡ The expected values are Ei j 216. The column totals are O 1 ¡ 214 and O2 121, O 3 The row totals are O1 total is O 430. ¡ 23. 244. The grand ¡ The null and alternate hypotheses are H0 : median of X 0. ...
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